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Is it true that a subspace M of a normed vector space X is closed if the limit of every sequence in M is contained in M? Whether or not X is complete? Are there alternative characterizations of closed?

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"...the limit of every convergent sequence in $M$ is contained in $M$."

Yes, this is one definition of a set being closed in a metric space.

You seem to be concerned with the following case: Suppose $X$ is not complete, and $\{v_n\}$ is a Cauchy sequence in $M$ that does not converge. Does this automatically mean $M$ is not closed, because the limit of $\{v_n\}$ is not in $M$? Well no, because $\{v_n\}$ doesn't have a limit. In order to conclude that $M$ is not closed, you would need to exhibit a sequence $\{v_n\}$ in $M$ that does converge, but whose limit is not in $M$.

Another characterization of closed subspaces: $M$ is closed iff $M^C$ is open; i.e. for every $v \in M^C$ there exists $r>0$ so that if $||v-w||<r$, $w \in M^C$ as well.

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  • $\begingroup$ It is necessary to be careful with the use of "convergent" ,as kccu points out. A Cauchy sequence in a metric space is not called a convergent sequence unless it has a limit point in that space. An infinite-dimensional normed vector space has vector sub-spaces on which the norm is not a complete metric. $\endgroup$ – DanielWainfleet Dec 6 '15 at 19:46
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That characterization is valid for arbitrary closed sets in any metric space.

On the one hand, closed sets are closed with respect to taking limits of sequences in arbitrary topological spaces. If $x$ is a limit of a sequence $(x_n)_n$ of points in a closed set $M$ then any neighbourhood of $x$ intersects the sequence and therefore $M.$

On the other hand, if $x$ belongs to the closure of a set $M$ in a metric space $X$ then any open ball $B(x,1/n)$ intersects $M$ so there exist points $x_n$ in $M$ that get arbitrarily close to $x.$

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